Description
You are given a sequence ofnintegersa1, a2, … , anin non-decreasing order. In addition to that, you are given several queries consisting of indicesiandj(1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integersai, … , aj.
Input
The input consists of several test cases. Each test case starts with a line containing two integersnandq(1 ≤ n, q ≤ 100000). The next line containsnintegersa1, … , an(-100000 ≤ ai≤ 100000, for eachi ∈ {1, …, n}) separated by spaces. You can assume that for eachi ∈ {1, …, n-1}: ai≤ ai+1. The followingqlines contain one query each, consisting of two integersiandj(1 ≤ i ≤ j ≤ n), which indicate the boundary indices for thequery.
The last test case is followed by a line containing a single0.
Output
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
Sample Input
10 3-1 -1 1 1 1 1 3 10 10 102 31 105 100
Sample Output
14
3
也许看完题你可能想到了线段树做法，但是这道题的确可以用RMQ来写。
具体就是查询区间中重复最多的次数，但是RMQ只能查询区间最大值，我们
初始化的时候也只能初始化连续的那一部分，如果区间中不连续了了。
所以我们要在查询上下功夫。从查询左端点起，找到不连续的点pos,pos
以后的就可以RMQ查询，，具体答案就是max(pos-l,RMQ(POS,l))
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<bitset>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair<int,int>pil;const int INF = 0x3f3f3f3f;const int maxn=1e5+100;int num[maxn],sum[maxn];int n,q;int dp[maxn][20];void init(){REPF(i,1,n)dp[i][0]=sum[i];for(int j=1;(1<<j)<=n;j++)for(int i=1;i+(1<<j)-1<=n;i++)dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int RMQ(int l,int r){int k=(int)(log(r-l+1)/log(2.0));return max(dp[l][k],dp[r-(1<<k)+1][k]);}void work(){int x,y,ans;while(q–){scanf("%d%d",&x,&y);int pos=x+1;while(num[pos]==num[pos-1]&&pos<=y) pos++;if(pos>y) ans=pos-x;else ans=max(pos-x,RMQ(pos,y));printf("%d\n",ans);}}int main(){int x;while(~scanf("%d",&n)&&n){scanf("%d",&q);CLEAR(sum,0);num[0]=INF;REPF(i,1,n){scanf("%d",&num[i]);if(num[i]==num[i-1])sum[i]=sum[i-1]+1;else sum[i]=1;}init();work();}return 0;}
没有什么可凭仗，只有他的好身体，没有地方可去，只想到处流浪。
